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3 years ago

10

Plz help......................

1 answer:

MaRussiya [10]3 years ago

8 0

Answer:

you'll get pushed back because of the energy force it happend to me before xD

Explanation:

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HELP THIS IS TIMED PLEASE OMG

Yakvenalex [24]

Answer:

d) 40000 kgm/s

Explanation:

The formula for momentum is p = m*v

so p = 100*400

p = 40000 kgm/s

7 0

3 years ago

A barge is 10 m wide and 60 m long and has vertical sides. The bottom of the boat is 1.2 m below the water surface. What is the

DENIUS [597]

Answer:

<h2>e. 7.1 MN approx.</h2>

Explanation:

Step one:

given data

density of water= 1000kg/m^3

the dimension of the barge

width= 10m

length= 60m

depth of the boat in the water= 1.2m

Hence the volume occupied by the boat is

volume=10*60*1.2

volume= 720m^2

Step two:

Required is the weight of the barge

we can first find the mass using the relation

density = mass/volume

mass= density*volume

mass= 1000*720

mass= 720000kg

Step three:

Weight =mg

g=9.81m/s^2

W=720000*9.81

W=7063200N

divided by 10^6

W=7.06MN

W=7.1MN approx.

5 0

3 years ago

Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

4 years ago

The paper dielectric in a paper-and-foil capacitor is 0.0800 mm thick. Its dielectric constant is 2.50, and its dielectric stren

jeka94

Answer:

a) 0.723 m²

b) 2000V

Explanation:

Given that

Thickness of the capacitor, d = 0.08*10^-3 m

Dielectric constant of the capacitor, k = 2.5

Dielectric strength of the capacitor, E = 50*10^6

Capacitance of the capacitor, C = 0.2*10^-6

Permittivity of free space, E• = 8.85*10^-12

a)

The area, A is given by the formula

A = (C * d) / (k * E•)

A = (0.2*10^-6 * 0.08*10^-3) / (2.5 * 8.85*10^-12)

A = 1.6*10^-11 / 2.213*10^-11

A = 0.723 m²

b)

Potencial difference, V is given by the formula

V = E * d

V = 1/2 * 50*10^6 * 0.08*10^-3

V = 1/2 * 4000

V = 2000 V

8 0

4 years ago

If the half-life of a 2.0 gram sample of a radionuclide is 15 hours, then the half-life of a 1.0 gram sample of the same radionu

Lerok [7]

7.5 hours or 450 minutes. 15/2=7.5

5 0

3 years ago